Question

Normal aluminum electrode coupled with normal hydrogen electrode gives an E.M.F. of 1.66 V. So the standard electrode potential of aluminum is: (Given that Galvanometer deflection is towards AI)

• A. -1.66 V
• B. 1.66 V
• C. -0.83 V
• D. +0.83 V

Answer 1

Answer: (B)

1.66 V

$2H^{+}+2e^{-}\to H_2{E}^0=0.0V$

${Al}^{3+}+3e^{-}\to Al{E}^0=?$

we know, $E_{cell}=E_{H^{+}/H_2}-{E^0}_{{Al}^{3+}/Al}$

$1.66V=0-{E^0}_{{Al}^{3+}/Al}$

${E^0}_{{Al}^{3+}/Al}=-1.66V$